Suppose we conduct a countably infinite number of independent Bernoulli trials, such that each trial results in $0$ w.p. $p$, or $1$ w.p. $q\equiv1-p$. What is the probability that the sequence $(1,0,1)$ occurs infinitely often?
My attempt: Let $(\Omega,\mathcal{F},\mathbb{P})$ be a suitable probability space. Assume $p\neq{0,1}$ (Otherwise, solution is obvious). Define:
$$A_n=\left\{\text{nth trial=1 ,(n+1)th trial=0, (n+2)th trial=1}\right\}$$
Then $\mathbb{P}(A_n)=pq^2$, and therefore:
$$\sum_{n=1}^{\infty}\mathbb{P}(A_n)=\infty$$
If the events $\left\{A_n\right\}_{n=1}^{\infty}$ were to be independent, then according to The Lemma of Borel Cantelli:
$$\mathbb{P}(A_n\text{ occurs infinitely often})=1$$
Problem is, I suspect the events $\left\{A_n\right\}_{n=1}^{\infty}$ are dependent. For example, if you look at $A_n$ and $A_{n+2}$, it seems like they are dependent. In general, $A_{m}$ and $A_{n}$ seem to be independent if $|m-n|\neq2$, but in the case where $|m-n|=2$ there seems to be a problem. I've been trying to find a workaronund, but to no avail.
Thanks!
Best Answer
The events $A_{3n}$ are independent and $\sum P(A_{3n})=\infty$. So $A_{3n}$'s occur infinitely many times with probability $1$. This implies that $A_n$'s occur infinitely many times with probability $1$.